Electronic – SPI routing question…Maximum length SPI can be routed

Rule-of-thumb says you have to calculate transmission line effects in if your connection length is longer than 1/10th of the signal's wavelength.

Transmission lines will cause reflections where they show a sudden change in impedance. The reflected signal adds to the original, may reflect again at the transmitter side, and that way go back and forth. The result is shown in the graph: the overshoot you're talking about, and some ringing.

A 12" wire length (30cm) is 1/10 of a 3m wavelength, or 100MHz. 1MHz, and even 4MHz shouldn't give much problems here. Terminating the line, like Wouter says, could be a first measure, though. The input is probably high impedance, and that's never a good matching. Since you have three different pieces in your connection (PCB, wire, PCB) finding the characteristic impedance will be difficult. (It will also be different for the PCB and the wire, so at very high frequencies you'd also get reflections at the cable's connections.) Given the short length and the low frequency the value of your terminating resistor is not critical. 100\$\Omega\$ may be too low for the microcontroller, most of them can't supply 33mA (or even 50mA, at 5V). I would try a 1k\$\Omega\$ resistor to start with, and see how far that gets us.

edit (re question update)
Rise time seems to be 7ns. That's fast, like Kortuk says, it means you have a spectrum at least to 400MHz, and those harmonics will indeed suffer transmission line effects, even if your clock is only 1MHz. Try to filter them out, a 20MHz bandwidth (80MHz for the 4MHz clock) gives you more than enough rise time. This is a 1MHz square wave filtered with a brick wall LPF at 20MHz:

Placing a series resistor will form a first order LPF with the line's capacitance. If we estimate that at 50pF then

\$ R = \dfrac{1}{2 \pi \cdot 100 MHz \cdot 50 pF} =32 \Omega \$

gives you cutoff frequency of 100MHz. So a 33\$\Omega\$ series resistor should decrease your rise time, but leave more than enough to have a good signal at 4MHz.

I'm afraid the answer is going to be a microcontroller or other programmable device. My reasoning is as follows:

1. The sensor is a slave device, not a master.

2. Slave devices need an external clock to clock the data out.

3. But only 24 clock pulses.

4. Simple logic devices aren't generally set up to make their own clock.

5. Therefore, the simplest solution will be a microcontroller with two SPI peripherals and sufficient memory to buffer through your maximum interrupt latency. Choose a micro with an internal clock and appropriate supply voltage for your circuit. Perhaps MSP430F5151.

Now, if your sensor was a SPI master, then you could be clever with multiplexed pins and just use a Microchip serial SRAM, setting up the write with the microcontroller, then letting the sensor continue writing into the RAM. See e.g., the Logic Shrimp logic analyzer for creative use of serial RAM parts.

Power consumption calculation with microcontroller.

By popular request, let's estimate the current draw for the MSP430 that I linked above.

MSP430 achieves low power with a switchable high-speed clock DCO, which will only need to run for the duration of the data transfer.

Let's choose 8 MHz SPI clock. 24-bit transfer will therefore take 24/(8 MHz) or 3 microseconds. From the datasheet, the DCO takes 1.6mA maximum and 6.5 microseconds maximum to start at 8 MHz. So the entire transfer takes about 10 microseconds. Each transfer takes 1.6 mA * 10 microseconds, or 16 nC. 8000 transfers / sec is 128 microamp average current. But double this, because we have to read out as well. (This is bit pessimistic, as the clock doesn't need to run during an SPI slave transaction.) So say an average current of 256 microamps.